SOLUTION: Someone, PLEASE HELP, this is a problem i have never seen before. i would appeciate any help just to get started. Evaluate P(-1/2) if P(x) = 2x4 + x3 + 12

Algebra ->  Expressions -> SOLUTION: Someone, PLEASE HELP, this is a problem i have never seen before. i would appeciate any help just to get started. Evaluate P(-1/2) if P(x) = 2x4 + x3 + 12      Log On


   



Question 143720: Someone, PLEASE HELP, this is a problem i have never seen before. i would appeciate any help just to get started.
Evaluate P(-1/2) if P(x) = 2x4 + x3 + 12

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Someone, PLEASE HELP, this is a problem i have never seen before. i would appeciate any help just to get started.
Evaluate P%28-1%2F2%29 if P%28x%29+=+2x%5E4+%2B+x%5E3+%2B+12
There are two methods to find P%28-1%2F2%29

First method:  by straight-forward plugging in -1%2F2 for x

P%28x%29+=+2x%5E4+%2B+x%5E3+%2B+12

P%281%2F2%29+=+2%28-1%2F2%29%5E4+%2B+%28-1%2F2%29%5E3+%2B+12

P%281%2F2%29+=+2%281%2F16%29+%2B+%28-1%2F8%29+%2B+12

P%281%2F2%29+=+1%2F8+-+1%2F8+%2B+12

P%281%2F2%29+=+12


Second method:  by synthetic division and the remainder
theorem which says that you get the same answer when you

1. plug a number directly into a polynomial 

as you get when you

2. write that number to the left of a synthetic division,
do the synthetic division, and take only the remainder,
which is the rightmost number on the bottom line of the
synthetic division. 

Notice that since the x%5E2 and x terms are 
missing in the original polynomial, we have to insert
0's for them as placeholders:

-1%2F2 | 2    1    0    0   12
     |     -1    0    0    0
      ----------------------
       2    0    0    0   12

The last number on the right 12 is the remainder
which by the remainder theorem is P%281%2F2%29


Edwin